使用wasapi绘制音频时遇到的问题

Question

过了一段时间，音频音高变得更清晰，然后只发出一些注释，这使我认为问题在于我填充缓冲区的方式

int AudioThreadProc(AUDIOSTRUCT* audio) {
    SetThreadPriority(GetCurrentThread(), THREAD_PRIORITY_HIGHEST);
    HANDLE buffReady = CreateEvent(0, 0, 1, 0);
    audio->client->lpVtbl->SetEventHandle(audio->client, buffReady);

    unsigned int soundBuffSize;
    audio->client->lpVtbl->GetBufferSize(audio->client, &soundBuffSize);
    audio->client->lpVtbl->Start(audio.client);

    static float time = 0;
    float dtime = 1 / (float)44100;

    for (;;) {
        WaitForSingleObject(buffReady, INFINITE);

        unsigned int soundFrames;
        audio->client->lpVtbl->GetCurrentPadding(audio->client, &soundFrames);
        unsigned int FramesToFill = soundBuffSize - soundFrames;
        unsigned short int* soundBuff;
        audio->render->lpVtbl->GetBuffer(audio.render, FramesToFill, &soundBuff);

        for (unsigned int i = 0; i < FramesToFill; i++) {
            unsigned short int soundWave = audio->callback(time);
            time += dtime;
            *soundBuff++ = soundWave;
            *soundBuff++ = soundWave;
        }
        audio->render->lpVtbl->ReleaseBuffer(audio.render, FramesToFill, 0);

    }
    return 0;
}

回调函数是唯一的正弦波发生器。

float callback(float time){
 returns sin(pitch * PI2 * time) * 0xFFFF;
}

//...
audio.callback = callback;

Answer 1

我认为这里的答案几乎相当于我几年前给出的this answer。随着time的上升，计算pitch*2π*time的浮点精度下降。调整您的代码，使time重置回零或is，并在完成完整的数据波后从零抵消。

就像这样：

time += dtime;
if (time > 1.0) {
    time -= 1.0;
}

此外，调用每个示例的函数调用的总开销也可能是一个问题。将callback转换为宏。你不会后悔的。